What is modal class? தமிழ் வாழ்க

Chemistry

2 answers:

german2 years ago

6 0

Answer:

Alenkasestr [34]2 years ago

6 0

Answer:

Modal means the one that occurs most often (averages: mode). In maths, the mode is one of the most common ways to describe a set of data. ... The modal class is, therefore, the group with the highest frequency. For example: if you counted the number of pencils in different pencil cases and you decided to group them.

Explanation:

தமிழன் ❤

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Describe the radial velocity curve. What is its shape? What is its amplitude? What is the orbital period?

lutik1710 [3]

Answer:

The radial velocity curve describes how fast a star is moving in its orbit around a center of mass ( m )

Curve amplitude : This is the maximum value of the radial velocity curve

Radial velocity shape ; The shape of Radial velocity curve is parabolic in nature

Orbital period : Orbital period is the time taken by the star to make one complete rotation in its orbit

Explanation:

The radial velocity curve describes how fast a star is moving in its orbit around a center of mass ( m ) while Curve amplitude is the maximum value of the radial velocity curve also The shape of Radial velocity curve is parabolic in nature. and Orbital period is the time taken by the star to make one complete rotation in its orbit

7 0

3 years ago

A solution made by dissolving 33 mg of insulin in 6.5 mL of water has an osmotic pressure of 15.5 mmHg at 25°C. Calculate the mo

Liula [17]

Answer: The molar mass of the insulin is 6087.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 15.5 mmHg

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 6.5 mL

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol$